Question

When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:Zn(s) + 2HCl(aq) --> ZnCl₂(aq) + H₂(g)

Answer 1

Answer : The enthalpy change per mole of zinc reacting for the reaction is 152.4 kJ/mol.

Explanation :

First we have to calculate the moles of Zn and HCl.

$\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}$

Molar mass of Zn = 65 g/mole

$\text{Moles of }Zn=\frac{1.34g}{65g/mole}=0.0206mole$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.750M\times 0.0600=0.0450mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

From the balanced reaction we conclude that

As, 1 mole of $Zn$ react with 2 mole of $HCl$

So, 0.0206 moles of $Zn$ react with $0.0206\times 2=0.0412$ moles of $HCl$

From this we conclude that, $HCl$ is an excess reagent because the given moles are greater than the required moles and $Zn$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the enthalpy change per mole of zinc reacting for the reaction.

From the reaction we conclude that,

As, 0.0206 moles of Zn produces heat = 3.14 kJ

So, 1 mole of Zn produces heat = $\frac{3.14kJ}{0.0206mol}=152.4kJ/mol$

Therefore, the enthalpy change per mole of zinc reacting for the reaction is 152.4 kJ/mol.